HW 3

Anastasiia Pryshchepa

1. Import the dataset Apartments.xlsx

library(readxl)
mydata <- read_excel("~/Desktop/Apartments.xlsx")
View(mydata)
mydata<-as.data.frame(mydata)
head(mydata)

##   Age Distance Price Parking Balcony
## 1   7       28  1640       0       1
## 2  18        1  2800       1       0
## 3   7       28  1660       0       0
## 4  28       29  1850       0       1
## 5  18       18  1640       1       1
## 6  28       12  1770       0       1

2. Change categorical variables into factors. (0.5 p)

mydata$ParkingF<-factor(mydata$Parking, levels = c(1,0), labels = c("Parking","No parking"))
mydata$BalconyF<-factor(mydata$Balcony, levels = c(1,0), labels = c("Balcony","No balcony"))
head(mydata)

##   Age Distance Price Parking Balcony   ParkingF   BalconyF
## 1   7       28  1640       0       1 No parking    Balcony
## 2  18        1  2800       1       0    Parking No balcony
## 3   7       28  1660       0       0 No parking No balcony
## 4  28       29  1850       0       1 No parking    Balcony
## 5  18       18  1640       1       1    Parking    Balcony
## 6  28       12  1770       0       1 No parking    Balcony

3. Test the hypothesis H0: Mu_Price = 1900 eur. What can you conclude? (1 p)

t.test(mydata$Price, mu = 1900, alternative = "two.sided")

## 
##  One Sample t-test
## 
## data:  mydata$Price
## t = 2.9022, df = 84, p-value = 0.004731
## alternative hypothesis: true mean is not equal to 1900
## 95 percent confidence interval:
##  1937.443 2100.440
## sample estimates:
## mean of x 
##  2018.941

#The mean is not equal to 1900, the mean of Price is 2018.941

4. Estimate the simple regression function: Price = f(Age). Save results in object fit1 and explain the estimate of regression coefficient, coefficient of correlation and coefficient of determination. (2 p)

fit1<-lm(Price~Age,data = mydata)
summary(fit1)

## 
## Call:
## lm(formula = Price ~ Age, data = mydata)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -623.9 -278.0  -69.8  243.5  776.1 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 2185.455     87.043  25.108   <2e-16 ***
## Age           -8.975      4.164  -2.156    0.034 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 369.9 on 83 degrees of freedom
## Multiple R-squared:  0.05302,    Adjusted R-squared:  0.04161 
## F-statistic: 4.647 on 1 and 83 DF,  p-value: 0.03401

#The estimate of the regression coefficient for the predictor variable “Age” is -8.975. This means that for each unit increase in the Age of an apartment, we expect the Price of the apartment to decrease by approximately $8.975, holding all other variables constant.Since p-value =0.034 (p<0.05), the regression coefficient is stattistically significant

#The coefficient of determination is 0.05302. This indicates that only 5.203% of total variability of Price can be explained by independant variable - Age. Since p-value<0.05, we can state that the coefficient of determination is greater than 0.

sqrt(summary(fit1)$r.squared)

## [1] 0.230255

#The linear correlation between age and price is direct and equals to 23.0255%

5. Show the scateerplot matrix between Price, Age and Distance. Based on the matrix determine if there is potential problem with multicolinearity. (0.5 p)

library(car)

## Loading required package: carData

scatterplotMatrix(mydata[c("Price","Age","Distance")], smooth = FALSE)

#No of the scateplots whosing the relationship between explanatory variables show strong relationship betweenn them, so it can be stated that there is no multicolinearity.

6. Estimate the multiple regression function: Price = f(Age, Distance). Save it in object named fit2.

fit2<-lm(Price~Age + Distance,data = mydata)
summary(fit2)

## 
## Call:
## lm(formula = Price ~ Age + Distance, data = mydata)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -603.23 -219.94  -85.68  211.31  689.58 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 2460.101     76.632   32.10  < 2e-16 ***
## Age           -7.934      3.225   -2.46    0.016 *  
## Distance     -20.667      2.748   -7.52 6.18e-11 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 286.3 on 82 degrees of freedom
## Multiple R-squared:  0.4396, Adjusted R-squared:  0.4259 
## F-statistic: 32.16 on 2 and 82 DF,  p-value: 4.896e-11

7. Chech the multicolinearity with VIF statistics. Explain the findings. (1 p)

vif(fit2)

##      Age Distance 
## 1.001845 1.001845

mean(vif(fit2))

## [1] 1.001845

#Since vif is close to 1, there is no strong correlation between explanatory variables, so no multicolinearity

8. Calculate standardized residuals and Cooks Distances for model fit2. Remove any potentially problematic units (outliers or units with high influence). (2 p)

mydata$StResid<-round(rstandard(fit2),3)
mydata$StResid

##  [1] -0.665  1.783 -0.594  0.754 -1.073 -0.778 -0.302 -0.787  0.455  1.241  1.551 -0.353 -1.499 -0.469  0.292 -1.000
## [17] -0.645  0.577 -0.812 -1.381 -0.371  1.576 -0.283 -0.440  1.571  0.628  1.550 -1.071 -0.516  0.256  0.889  0.454
## [33]  2.051 -0.663 -1.261 -1.189 -1.005  2.577  1.091  0.279 -1.052 -0.103  0.252  0.117 -0.329  1.478 -0.053 -1.009
## [49]  0.747 -0.683  0.393 -0.861 -2.152 -1.101  1.445 -0.328  1.601  1.655 -0.550  0.333  1.783 -0.594  0.754 -1.073
## [65] -0.778 -0.302 -0.787  0.455  1.241  1.551 -0.353 -1.499 -0.469  0.292 -1.000 -0.645 -0.329  1.478 -0.053 -1.009
## [81]  0.747 -0.683  0.393 -0.861 -0.053

mydata$CooksD<-round(cooks.distance(fit2),3)
mydata$CooksD

##  [1] 0.007 0.030 0.006 0.008 0.005 0.005 0.001 0.004 0.001 0.017 0.020 0.001 0.013 0.002 0.001 0.006 0.004 0.003 0.010
## [20] 0.012 0.002 0.061 0.001 0.001 0.034 0.004 0.023 0.005 0.003 0.001 0.030 0.002 0.069 0.005 0.008 0.012 0.007 0.320
## [39] 0.038 0.002 0.005 0.000 0.001 0.000 0.003 0.019 0.000 0.022 0.004 0.003 0.004 0.003 0.066 0.012 0.104 0.000 0.032
## [58] 0.037 0.001 0.001 0.030 0.006 0.008 0.005 0.005 0.001 0.004 0.001 0.017 0.020 0.001 0.013 0.002 0.001 0.006 0.004
## [77] 0.003 0.019 0.000 0.022 0.004 0.003 0.004 0.003 0.000

head(mydata[order(-mydata$CooksD), c("Price","CooksD")], 6)

##    Price CooksD
## 38  2180  0.320
## 55  1740  0.104
## 33  2790  0.069
## 53  1760  0.066
## 22  2540  0.061
## 39  2400  0.038

library(dplyr)

## 
## Attaching package: 'dplyr'

## The following object is masked from 'package:car':
## 
##     recode

## The following objects are masked from 'package:stats':
## 
##     filter, lag

## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union

mydata_0<-mydata %>%
  filter(!Price %in% c(2180,1740,2790,1760,2540,2400))

9. Check for potential heteroskedasticity with scatterplot between standarized residuals and standrdized fitted values. Explain the findings. (1 p)

mydata$StdFittedValues<-scale(fit2$fitted.values)
library(car)
scatterplot(y=mydata$StResid, x=mydata$StdFittedValues, ylab = "Standardized residuals", xlab = "Standardized fitted values", boxplots = FALSE, regLine = FALSE, smooth = FALSE)

#There is no heteroskedasticity

10. Are standardized residuals ditributed normally? Show the graph and formally test it. Explain the findings. (1 p)

mydata$StResid<-round(rstandard(fit2),3)
hist(mydata$StResid, xlab = "Standardized residuals", ylab = "Frequency", main = "Histogram of standardized residuals")

shapiro.test(mydata$StResid)

## 
##  Shapiro-Wilk normality test
## 
## data:  mydata$StResid
## W = 0.95303, p-value = 0.003645

#Standard residuals are not normally distributed

11. Estimate the fit2 again without potentially excluded units and show the summary of the model. Explain all coefficients. (2 p)

fit2<-lm(Price~Age + Distance,data = mydata)
summary(fit2)

## 
## Call:
## lm(formula = Price ~ Age + Distance, data = mydata)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -603.23 -219.94  -85.68  211.31  689.58 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 2460.101     76.632   32.10  < 2e-16 ***
## Age           -7.934      3.225   -2.46    0.016 *  
## Distance     -20.667      2.748   -7.52 6.18e-11 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 286.3 on 82 degrees of freedom
## Multiple R-squared:  0.4396, Adjusted R-squared:  0.4259 
## F-statistic: 32.16 on 2 and 82 DF,  p-value: 4.896e-11

#The estimate of the regression coefficient for the predictor variable “Age” is -9.793. This means that for each unit increase in the Age of an apartment, we expect the Price of the apartment to decrease by approximately $9.793, holding all other variables constant.Since p-value <0.05, the regression coefficient is stattistically significant

#The estimate of the regression coefficient for the predictor variable “Age” is -23.604. This means that for each unit increase in the Age of an apartment, we expect the Price of the apartment to decrease by approximately $23.604, holding all other variables constant.Since p-value <0.05, the regression coefficient is stattistically significant

#The coefficient of determination is 0.5382. This indicates that only 53.82% of total variability of Price can be explained by independant variables - Age and Distance. Since p-value<0.05, we can state that the coefficient of determination is greater than 0.

sqrt(summary(fit2)$r.squared)

## [1] 0.6629887

#The linear correlation between independant variables and dependant ones is direct and equals to 73.366458%

12. Estimate the linear regression function Price = f(Age, Distance, Parking and Balcony). Be careful to correctly include categorical variables. Save the object named fit3.

fit3<-lm(Price~Age + Distance + ParkingF + BalconyF,data = mydata)
summary(fit3)

## 
## Call:
## lm(formula = Price ~ Age + Distance + ParkingF + BalconyF, data = mydata)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -459.92 -200.66  -57.48  260.08  594.37 
## 
## Coefficients:
##                    Estimate Std. Error t value Pr(>|t|)    
## (Intercept)        2499.770     80.645  30.997  < 2e-16 ***
## Age                  -6.799      3.110  -2.186  0.03172 *  
## Distance            -18.045      2.758  -6.543 5.28e-09 ***
## ParkingFNo parking -196.168     62.868  -3.120  0.00251 ** 
## BalconyFNo balcony   -1.935     60.014  -0.032  0.97436    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 273.7 on 80 degrees of freedom
## Multiple R-squared:  0.5004, Adjusted R-squared:  0.4754 
## F-statistic: 20.03 on 4 and 80 DF,  p-value: 1.849e-11

13. With function anova check if model fit3 fits data better than model fit2. (1 p)

anova(fit2,fit3)

## Analysis of Variance Table
## 
## Model 1: Price ~ Age + Distance
## Model 2: Price ~ Age + Distance + ParkingF + BalconyF
##   Res.Df     RSS Df Sum of Sq      F  Pr(>F)  
## 1     82 6720983                              
## 2     80 5991088  2    729894 4.8732 0.01007 *
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

#Model2 improves the fit compared to Model 1 by the decrease in RSS. Withe the confidence level of 90%, we can state that the second model fits the data better (p<0.1)

14. Show the results of fit3 and explain regression coefficient for both categorical variables. Can you write down the hypothesis which is being tested with F-statistics, shown at the bottom of the output? (2 p)

summary(fit3)

## 
## Call:
## lm(formula = Price ~ Age + Distance + ParkingF + BalconyF, data = mydata)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -459.92 -200.66  -57.48  260.08  594.37 
## 
## Coefficients:
##                    Estimate Std. Error t value Pr(>|t|)    
## (Intercept)        2499.770     80.645  30.997  < 2e-16 ***
## Age                  -6.799      3.110  -2.186  0.03172 *  
## Distance            -18.045      2.758  -6.543 5.28e-09 ***
## ParkingFNo parking -196.168     62.868  -3.120  0.00251 ** 
## BalconyFNo balcony   -1.935     60.014  -0.032  0.97436    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 273.7 on 80 degrees of freedom
## Multiple R-squared:  0.5004, Adjusted R-squared:  0.4754 
## F-statistic: 20.03 on 4 and 80 DF,  p-value: 1.849e-11

#Apartments with no parking facilities tend to have by $142.085 lower prices compared to those with parking facilities.

#Apartments with no balcony tend to have by $7.028 higher prices than those with